The combined set (linear form)

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A known set of the set B, and the first set of data within one unique. Seeking A, B set into a new set of C, C required in the data set are unique. And pointed out that the number of C collection.

Input
three lines, the first line A are set, the number B of
data A second set of behavioral
data of the third set of behavior B
output of
two lines of
a first row number set of the C
data of the second set of behaviors C
Sample input
. 4. 5
12 is 34 is 56 is 78
34 is 67 89 34 is 76
sample output
. 7
12 is 34 is 56 is 78 67 89 76
tips
data is less than 30,000

AC Code:

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=10000;
template<class T>
struct node
{
   int data;
   //node<T> *r;
    node<T>* next;
};
template<class T>
class lian
{
   node<T> *first;
   node<T> *r;
   public:
   lian();
   lian(T a[],int n);
   void cha(int a);
   void show();
   //void zhishan(T n);
   //void weishan(int n);
};
template <class T>
lian<T>::lian()
{
    first=new node<T>;
    first->next=NULL;
}
template <class T>
lian<T>::lian(T a[],int n)
{
    first=new node<T>;
    //node<T> *r=first;
    r=first;
     node<T> *s=NULL;
    for(int i=0;i<n;i++)
    {
        s=new node<T>;
        s->data=a[i];
        r->next=s;
        r=s;
    }
    r->next=NULL;
}
template <class T>
void lian<T>::cha(int a)
{
    node<T> *s;
    s=new node<T>;
    s->data=a;
     node<T> *p;
    p=first->next;
    while(p!=NULL)
    {
       if(p->data==a)return;
        p=p->next;
    }
    r->next=s;
    r=s;
    r->next=NULL;
}
template <class T>
void lian<T>::show()
{
    node<T> *p;
    p=first->next;
    int t=0,h[maxn];
    while(p!=NULL)
    {
        h[t]=p->data;
        p=p->next;
        t++;
    }
    cout<<t<<endl;
    for(int i=0;i<t;i++)
    {
        cout<<h[i]<<" ";
    }
    cout<<endl;
}
int main()
{
   int a[maxn],b[maxn];
   int x,y;
   cin>>x>>y;
   for(int i=0;i<x;i++)
   {
       cin>>a[i];
   }
   for(int i=0;i<y;i++)
   {
       cin>>b[i];
   }
   lian<int> k(a,x);
 for(int i=0;i<y;i++)
 {
     k.cha(b[i]);
 }
   k.show();

}